For a reaction $A + B \rightarrow \text{product}$,if $[A]$ is doubled keeping $[B]$ constant,the rate of reaction doubles. Calculate the order of reaction with respect to $A$.

For the reaction $2N_2O_5 \to 4NO_2 + O_2$,the rate of reaction and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively. The concentration of $N_2O_5$ at that time will be:

For the reaction $2A + B \rightarrow A_2B$,if the concentration of reactant $A$ is doubled and the concentration of reactant $B$ is halved,the rate of the reaction will:

The value of $\frac{t_{0.875}}{t_{0.50}}$ for $n^{th}$ order reaction is

The rate law for a reaction between the substances $A$ and $B$ is given by,$rate = k[A]^n[B]^m$. On doubling the concentration of $A$ and halving the concentration of $B$,the ratio of the new rate to the earlier rate of the reaction will be as

Inversion of cane sugar is an example of $-$